Two circles $C_{1}$ and $C_{2}$ intersect at points $A$ and $B$. Let $P$, $Q$ be points on circles $C_{1}$, $C_{2}$ respectively, such that $|AP| = |AQ|$. The segment $PQ$ intersects circles $C_{1}$ and $C_{2}$ in points $M$, $N$ respectively. Let $C$ be the center of the arc $BP$ of $C_{1}$ which does not contain point $A$ and let $D$ be the center of arc $BQ$ of $C_{2}$ which does not contain point $A$ Let $E$ be the intersection of $CM$ and $DN$. Prove that $AE$ is perpendicular to $CD$. Proposed by Steve Dinh
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Tags: geometry, emc
Ankoganit
01.01.2017 07:17
[asy][asy]size(7.5cm);pair A=(3.46,4.04), B=(2.79,-0.35), P=(0.39,6.05), Q=(4.73,0.59), M=(3.84,1.71), C=(-3.40,0.98), D=(3.88,-0.13),Nn=(2.06,3.95); pair Ee=extension(C,M,D,Nn); draw(circumcircle(A,B,C)^^circumcircle(A,B,D),green); D(MP("A",A,NE)--MP("P",P,N)--MP("N",Nn,NE)--MP("M",M,E)--MP("Q",Q,SE)--A,heavymagenta); D(A--MP("C",C,SW)--MP("D",D,S)--A--MP("B",B,S),heavycyan); draw(D--Nn^^C--M,darkmagenta+dashed); dot(A);dot(B);dot(C);dot(D);dot(P);dot(Q);dot(M);dot(Nn);D(MP("E",Ee,SW),black); [/asy][/asy] Okay, let's chase down some angles. \begin{align*}\angle CAD &=\angle CAB+\angle BAD \\ &=\frac 12\angle PAB+\frac 12\angle BAQ\\ &=\frac 12\angle PAQ=90^{\circ}-\angle AQN \text{ (since }\triangle APQ\text{ is }A-\text{isosceles)}\\ &= 90^{\circ}-\angle ADN\\ \implies \angle CAD &= 90^{\circ}-\angle ADN\end{align*}This implies $DN\perp AC$; similarly $CM\perp AD$, so $E$ is in fact the orthocenter of $\triangle ACD$. The fact that $AE\perp CD$ follows easily. $\blacksquare$
Mahdi_Mashayekhi
25.12.2021 17:50
Easy one we will prove E is orthocenter of ACD. ∠PAC = ∠CAB and ∠BAD = ∠DAQ and PAQ is isosceles so ∠CAD + ∠AQP = 90. ∠CAD + ∠AQP = ∠CAD + ∠ADN = 90 ---> DE ⊥ AC. ∠CAD + ∠APQ = ∠CAD + ∠ACM = 90 ---> CE ⊥ AD. so E is orthocenter and we're Done.